Of course. I will show it on numbers in $[0,1)$ and $[0,1)\times[0,1)$. Consider $z=x+iy$ with $x=0.x_1x_2x_3\ldots$ and $y=0.y_1y_2y_3\ldots$ their decimal expansions (the standard, greedy ones with no $9^\omega$ as a suffix). Then the number $f(z)=0.x_1y_1x_2y_2x_3y_3\ldots$ is real and this map is clearly injective on the above mentioned sets. Generalization to the whole $\mathbb C$ is straighforward. This gives $\#\mathbb C\leq\#\mathbb R$. the other way around is obvious.

This requires a bit more work. The map isn’t well-defined until you deal with the $0.4999\dots=0.5000\dots$ issue; if you deal with that straightforwardly, it’a nor surjective.
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Brian M. ScottNov 26 '12 at 19:14

Yes, you are right. However, they all all (complex) rational hence of no interest for the sets of continuum cardinality. I'll add a comment.
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yo'Nov 26 '12 at 19:16

And btw, usually a string with suffix $9^\omega$ is not considered to be an expansion (it is only a representation), in the usual greedy expansions as defined by Rényi in 1957.
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yo'Nov 26 '12 at 19:22

I’ve never seen anyone make a distinction between representation and expansion, and I very much doubt that the distinction can be considered standard; certainly it does not qualify as well-known, so if you use it, you need to explain it.
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Brian M. ScottNov 26 '12 at 19:29

1

And yes, I know that only countably many numbers are affected and that this does not affect the result, but I don’t know that the OP knows this.
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Brian M. ScottNov 26 '12 at 19:33